I would like to draw your attention to appendix $C$ on page 38 of this paper.

The equation $C.2$ there seems to be evaluating the sum $\sum_R \chi _R (U^m)$ in equation 3.16 of this paper. I understand that $U \in U(N_c)$ group and then with $N_f$ (fundamental?) hypermultiplets it seems that apparently the following is being claimed to be true,

I was pretty much confused with the idea in equation $C.5$ and $C.6$ of thinking the eigenvalues of the $U(N_c)$ group on the circle to be given by a distribution function $\rho (\theta) \geq 0$ with a mysterious normalization as in equation $C.5$.

I would be glad if someone can help understand how equation C.6 was determined.
A lot of steps seem to be skipped here which I found hard to reinstate!

Going by this paper it seems that the original more general equation $2.43$ on page 17 has two possible limits as in equation $2.45$ on page 17 when there are $c$ matter fields in the adjoint and $N_c \rightarrow \infty$ or equation $2.48$ in the Veneziano limit with fundamental matter fields.

I would like to know the derivation/reference for the equations 2.45 and 2.48 cited above.

I guess this is one of the most educative means of taking the Veneziano limit in which AdS/CFT is supposed to work?

1 Answer
1

The character $\chi_R : G \to \mathbb{C}$ of a representation $R$ is defined by $\chi_R(U) = Tr_R(U)$, namely by taking the trace of $U$ in the representation $R$. See for example Appendix A of Aharony et al.. Then the equation you write seems reasonable: presumably there are $N_f$ hypermultiplets, each with fundamental and anti-fundamental fields that give $Tr(U^m)$ and $Tr(U^{-m})$
respectively in the sum over $R$. The trace is taken of the matrix $U$ because the fields are in the fundamental.

The factor of 2 should come from the details of which representations exactly are being summed over, the field content of the hypermultiplets, etc., which I didn't try to follow.

Now, regarding the distribution of eigenvalues that appears in (C.5), let me give you a physical explanation of where this comes from. They consider the thermodynamics of a theory on a sphere, so the topology is that of a sphere times $S^1$ -- the Euclidean compact time. In this case there is a zero-mode of the gauge field, which comes from the fact that you can turn on a constant field in the thermal direction. This mode cannot be removed by a gauge transformation (in general), so in a path integral you need to integrate over it. This is explained for example in section 4.1 of Aharony et al. You can use a gauge transformation to diagonalize this zero-mode matrix, so you are left with the discrete eigenvalues of the matrix to integrate over. You can also show that these eigenvalues live on a circle, because you can shift them by $2\pi$ (in some normalization) using a gauge transformation, so you need to integrate them over a circle. In the large $N$ limit you have an infinite number of such eigenvalues, but they are still restricted to live on a circle. So you need to describe them using a density, and that is what they call the distribution $\rho$.

When taking $N \to \infty$ the delta functions in $\rho$ will become very dense, and we will be able to approximate $\rho$ by a smooth function, whose value at $\theta$ depends on the density of the delta functions across the small interval $[\theta,\theta+\epsilon]$. This will be a good approximation because the function $\cos(n\theta)$ that we are integrating over will not vary much in these intervals, so averaging it instead of sampling it will not change the result by much. When we reach the limit $N=\infty$ this will no longer be an approximation, because the delta functions will become continuous.

So this explains the derivation of the second term in (C.6). As for the first term, the idea is similar except that you have two integrals over $\theta,\theta'$ corresponding to the $i,j$ sums. The contributions from the $i=j$ terms ($\theta=\theta'$ in the integral) are subleading in the large $N$ limit: they scale as $N$ while the rest of it scales as $N^2$, so they are neglected.

Now, replace $\cos(n(\theta-\theta')) = \cos(n\theta) \cos(n\theta') + \sin(n\theta) \sin(n \theta')$. Notice that in the path integral over $\rho$ you may consider only eigenvalue distributions that are symmetric under $\theta \to -\theta$, because the original integrand is invariant under this. This is explicitly mentioned for example in Schnitzer, who does the same computation. So this means that $\int d\theta \rho \sin(n\theta) = 0$. I think it should be clear from this point.

@ Guy Gur-Ari Each of the $N_f$ hypermultiplets I think has 2 $N=2$ chiral multiplets one in the adjoint and the other in the conjuagte adjoint (= adjoint) of the gauge group. Now we have $\chi_{adj} = \chi_{\bar{adj}} = \chi_{fund}\times \chi_{anti-fund}$. So each hypermultiplet should have contributed a $2\chi_{fund}\times \chi_{anti-fund}$..right?
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user6818Jul 31 '12 at 21:57

I can see this general argument of needing to describe the eigenvalue of a $U(N)$ matrix by a density on the circle buts its the specific re-writing of that as in equation C.6 that is puzzling me. It would be a great help if you can may be write in a few lines about how that equation was gotten!
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user6818Jul 31 '12 at 22:00

They matter multiplets they consider in appendix C are in the fundamental, as they write in the beginning of section 3.2 (a bit before they refer to the appendix). I added the explanation of (C.6) to the answer.
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Guy Gur-AriAug 1 '12 at 0:16

Thanks for the updates! I think I am missing something here. In the beginning of section 2.2 (page 7) don't they say that the 2 N=2 chiral multiplets inside the N=3 hypermultiplet are in conjugate representations of the gauge group? Then is it wrong to think that these 2 N=2 chiral components are in $Adj = \bar{Adj}$ of the gauge group? If the $N_f$ hypermultiplets are in the fundamental then what is the representation in which the 2 $N=2$ chiral multiplets fall into? It will be great if you can clarify this point!
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user6818Aug 1 '12 at 18:47

As they say, in the $N=3$ theory you can have hypermultiplets, and each hypermultiplet has two chiral multiplets: one in a rep $R$ and one in $\bar{R}$. It is not necessary that $R$ is the adjoint. A bit below the place you mention they consider various possibilities for $R$, only one of which is the adjoint. If you say that a hypermultiplet is 'fundamental' (which I think is not a precise statement although its meaning is clear), it means that you have one chiral multiplet in the fundamental and one in the anti-fundamental.
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Guy Gur-AriAug 1 '12 at 19:10